Question

White light (400 to 700 nm) is used to illuminate a double slit with a spacing of 1.25 mm. An interference pattern is observed on a screen placed 1.5 m away. A small hole located 3 mm above the white central stripe allows a small portion of the pattern to be analyzed by a high resolution spectrograph. What wavelengths will be absent in the portion of light analyzed by the spectograph?

Answer 1

Let us calculate the distance of the 1st maxima of all the wavelengths,

d sin (theta) = n × lambda

Y = (n lambda) × (D/d)

Putting n = 1 and finding the range of Y from lambda = 400 nm to lambda = 700 nm

Y = 0.48 mm to 0.84 mm

First minima from 0.24 mm to 0.42 mm

So at 3 mm distance

0.375 mm, 0.333 mm, 0.3mm, 0.272mm, 0.25 will be having their minima ( as they all divide 3mm)

Let us calculate the corresponding wavelengths

By d sin(theta) =(n +1/2) (lambda) , n starts from zero

625 nm, 555nm, 500nm, 453.33nm, 416.67nm

These wavelengths will be absent